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      【二叉树】 给出二叉树的中序遍历 和先序遍历或后序遍历 确定二叉树.md
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        <h2 id="说明-2022-05-05"><a class="markdownIt-Anchor" href="#说明-2022-05-05"></a> 说明 - 2022-05-05</h2>
<p>本篇博客为本人原创, 原发布于CSDN, 在搭建个人博客后使用爬虫批量爬取并挂到个人博客, 出于一些技术原因博客未能完全还原到初始版本(而且我懒得修改), 在观看体验上会有一些瑕疵 ,若有需求会发布重制版总结性新博客。发布时间统一定为1111年11月11日。钦此。</p>
<p><strong>讨论的内容如题</strong></p>
<h2 id="分析"><a class="markdownIt-Anchor" href="#分析"></a> 分析</h2>
<p>拿中序遍历和后序遍历的组合举例子<br />
设数组last[]储存后续遍历 in[]中序遍历，它们的长度都是n(数组从1开始计，下标范围是1-n)</p>
<p>树根自然为last[n]<br />
，这时我们遍历in[]数组，在in[]数组中找到last[n]的值的下标idx，因为in[]储存的是中序遍历，显然我们以idx为界把in数组的序列分成两份[1,idx-1],[idx+1,n]，分别为两颗子树上的节点。</p>
<p>之后我们便可以在分出的两个区间之中重复上述操作，分别找目前子树的根节点，再二分区间，直到获得完整的二叉树</p>
<p>可以使用递归来完成 函数(区间左，区间右，根节点，偏移量)</p>
<p>此时注意一个问题：当区间被二分之后，我们发现后面的区间在in[]数组和last[]数组中的下标“对不齐”了，即下标发生了偏移，如下图。所以要引入一个变量shift来记录偏移量。（偏移量最开始是0，递归时前面的区间继承原偏移量，后面的区间偏移量为原偏移量+1）<br />
![在这里插入图片描述](<a target="_blank" rel="noopener" href="https://img-blog.csdnimg.cn/20201008231828150.png?x-oss-">https://img-blog.csdnimg.cn/20201008231828150.png?x-oss-</a><br />
process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dsZGNtenk=,size_16,color_FFFFFF,t_70#pic_center)<br />
（在前序遍历和中序遍历的组合中，下标发生偏移的是前面的区间）</p>
<h2 id="例题-后序遍历和中序遍历的组合"><a class="markdownIt-Anchor" href="#例题-后序遍历和中序遍历的组合"></a> 例题 后序遍历和中序遍历的组合</h2>
<p><a target="_blank" rel="noopener" href="https://vjudge.net/contest/386054#problem/E">https://vjudge.net/contest/386054#problem/E</a></p>
<p>​<br />
#include <iostream><br />
#include <cstring><br />
#include <cstdio><br />
struct Node<br />
{<br />
int l,r;<br />
};<br />
Node treeN[10010];<br />
int inorder[10010],postorder[10010];<br />
int len,rt,least,lstPos;<br />
int input()//返回数组长度<br />
{<br />
int n,idx=1;<br />
char ch;<br />
while(scanf(&quot;%d&quot;,&amp;n)!=EOF)<br />
{<br />
inorder[idx++]=n;<br />
ch=getchar();<br />
if(ch==’\n’)<br />
break;<br />
}<br />
idx=1;<br />
while(scanf(&quot;%d&quot;,&amp;n)!=EOF)<br />
{<br />
postorder[idx++]=n;<br />
ch=getchar();<br />
if(ch==’\n’)<br />
break;<br />
}<br />
return idx-1;<br />
}<br />
int makeTree(int l,int r,int root,int n)<br />
{<br />
if(l&gt;r)<br />
return -1;<br />
int idx=-1;<br />
for(int i=l; i&lt;=r; i++)<br />
{<br />
if(inorder[i]<mark>root)<br />
{<br />
idx=i;<br />
break;<br />
}<br />
}<br />
if(idx&gt;0)<br />
{<br />
treeN[root].l=makeTree(l,idx-1,postorder[idx-n-1],n);<br />
treeN[root].r=makeTree(idx+1,r,postorder[r-n-1],n+1);<br />
}<br />
return root;<br />
}<br />
void add(int root,int base)<br />
{<br />
int total=base+root;<br />
if(treeN[root].l</mark>-1 &amp;&amp; treeN[root].r==-1)<br />
{<br />
if(lstPos==-1)<br />
{<br />
least=total;<br />
lstPos=root;<br />
}<br />
else<br />
{<br />
if(total&lt;least)<br />
{<br />
least=total;<br />
lstPos=root;<br />
}<br />
}<br />
return ;<br />
}<br />
if(treeN[root].l&gt;0)<br />
{<br />
add(treeN[root].l,total);<br />
}<br />
if(treeN[root].r&gt;0)<br />
{<br />
add(treeN[root].r,total);<br />
}<br />
}</p>
<pre><code>int main()
&#123;
    while((len=input())&gt;0)
    &#123;
        memset(treeN,-1,sizeof(treeN));
        lstPos=-1;
        rt=makeTree(1,len,postorder[len],0);
        add(rt,0);
        std::cout &lt;&lt; lstPos &lt;&lt; std::endl;
    &#125;
    return 0;
&#125;
</code></pre>
<p>​</p>
<h2 id="例题-前序遍历和中序遍历的组合"><a class="markdownIt-Anchor" href="#例题-前序遍历和中序遍历的组合"></a> 例题 前序遍历和中序遍历的组合</h2>
<p>&lt;<a target="_blank" rel="noopener" href="https://pintia.cn/problem-">https://pintia.cn/problem-</a><br />
sets/1303281293857476608/problems/1303281404742291466&gt;</p>
<p>​<br />
#include <iostream><br />
#include <cstdio><br />
#include <cstring><br />
#include <algorithm><br />
#include <iomanip><br />
#include <queue><br />
#include <cmath><br />
#include <map></p>
<pre><code>const int N=32;
int n;
struct Node
&#123;
    int l,r;
&#125;treeN[N];
int pre[N],in[N];

int fff(int root,int l,int r,int shift)
&#123;
    if(l&gt;r) return -1;
    int i;
    for(i=l; i&lt;=r; i++) if(root==in[i]) break;
    if(i&lt;=r)
    &#123;
        treeN[root].l=fff(pre[l+1+shift],l,i-1,shift+1);
        treeN[root].r=fff(pre[i+1+shift],i+1,r,shift);
    &#125;
    return root;
&#125;

void rev(int root)
&#123;
    if(root==0) return ;
    treeN[root].r=treeN[root].r^treeN[root].l;
    treeN[root].l=treeN[root].r^treeN[root].l;
    treeN[root].r=treeN[root].r^treeN[root].l;
    rev(treeN[root].l);
    rev(treeN[root].r);
&#125;

void gogogo(int root)
&#123;
    std::queue&lt;int &gt;q;
    std::cout &lt;&lt; root;
    if(treeN[root].l!=-1) q.push(treeN[root].l);
    if(treeN[root].r!=-1) q.push(treeN[root].r);
    while(!q.empty())
    &#123;
        root=q.front(); q.pop();
        std::cout &lt;&lt; &quot; &quot; &lt;&lt; root;
        if(treeN[root].l!=-1) q.push(treeN[root].l);
        if(treeN[root].r!=-1) q.push(treeN[root].r);
    &#125;
&#125;

int main()
&#123;
    std::cin &gt;&gt; n;
    for(int i=0; i&lt;n; i++) std::cin &gt;&gt; in[i];
    for(int i=0; i&lt;n; i++) std::cin &gt;&gt; pre[i];
    int root=fff(pre[0],0,n-1,0);
    rev(root);
    gogogo(root);
    return 0;
&#125;
</code></pre>
<p>​</p>

      
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